Last Updated :
29 Oct, 2018
Consider the following statements regarding page table and inverted page table.
- A Computer System with 48-bit logical address space and 1Kbytes page size. The system has 1 Gbytes of physical memory. This implies that there are 512 K entries in an inverted page table.
- A Computer System with 32-bit logical address space and 4Kbytes page size. The system has 1 Gbytes of physical memory and page table entries size is 4 bytes. This implies that the size of page table is 2MB.
Which of the following option is correct?
(A) only statement 1 is correct.
(B) both statement 1 and 2 are correct
(C) only statement 2 is correct
(D) neither statement 1 nor 2 is correct
Answer: (D)
Explanation:
- Number of entries in inverted page table = physical address space / page size = 230/10 = 20 = 1024 K entries.
- Since we have a virtual address space of 2^32 and each page size is 2^12, we can store (2^32/2^12) = 2^20 pages. Since each entry into this page table has an address of size 4 bytes, then we have 2^20*4 = 4MB. So the page table takes up 4MB in memory.
So, neither statement 1 nor 2 is correct.
Quiz of this Question
Share your thoughts in the comments
Please Login to comment...